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对任意正整数 ${\displaystyle n}$, 记集合 $$\begin{array}{rl}{A}_n& =\{(a_1,\cdots ,a_n)|a_1,\cdots ,a_n\text{均为非负整数,且}{a}_1+\cdots +a_n=n\},\\ B_n& =\{(b_1,\cdots ,b_n)|b_1,\cdots ,b_n\text{均为非负整数,且}{b}_1+\cdots +b_n=2n\}.\end{array}$$ 设 ${\displaystyle \alpha =(a_1,\cdots ,a_n)\in A_n}$, ${\displaystyle \beta =(b_1,\cdots ,b_n)\in B_n}$. 若对任意 ${\displaystyle i\in \{1,\cdots ,n\}}$ 都有 ${\displaystyle a_i\le b_i}$, 则称 ${\displaystyle \alpha \le \beta }$.

1. 写出集合 ${\displaystyle A_2}$ 和 ${\displaystyle B_2}$;
2. 取 ${\displaystyle {\alpha }_3\in A_3}$, ${\displaystyle {\alpha }_3=(1,0,2)}$, 写出两个 ${\displaystyle B_3}$ 中的元素 ${\displaystyle {\beta }_3,{\beta }_4}$, 使得 ${\displaystyle {\alpha }_3\le {\beta }_i}$ (${\displaystyle i=3,4}$);
3. 证明: 对任意 ${\displaystyle \alpha \in A_n}$, 存在 ${\displaystyle \beta \in B_n}$, 使得 ${\displaystyle \alpha \le \beta }$.

1. 解 ${\displaystyle A_2=\{(0,2),(1,1),(2,0)\}}$, ${\displaystyle B_2=\{(0,4),(1,3),(2,2),(3,1),(4,0)\}}$.
2. 解 可以取 ${\displaystyle {\beta }_3=(1,2,3)}$, ${\displaystyle {\beta }_4=(2,2,2)}$.
3. 证明 记 ${\displaystyle \alpha =(a_1,\cdots ,a_n)}$, 我们直接取 ${\displaystyle \beta =(a_1+1,\cdots ,a_n+1)}$. 容易验证 ${\displaystyle \beta \in B_n}$, 且 ${\displaystyle \beta \ge \alpha }$.